Question: Solve for $x$: $4x^{1/3}-2 \cdot \frac{x}{x^{2/3}}=7+\sqrt[3]{x}$.
To start, note that $\frac{x}{x^{2/3}}=x^{1-\frac{2}{3}}=x^{1/3}$.  Also note that we can rewrite the cube root with a fractional exponent, so $\sqrt[3]{x}=x^{1/3}$.  Using these pieces of information, rewrite the given equation as: $$4x^{1/3}-2x^{1/3}=7+x^{1/3}$$ Move all $x^{1/3}$ terms to one side and simplify: \begin{align*}
2x^{1/3}-x^{1/3}&=7\\
\Rightarrow\qquad x^{1/3}&=7\\
\Rightarrow\qquad (x^{1/3})^3&=7^3\\
\Rightarrow\qquad x&=\boxed{343}
\end{align*}